Pointwise convergence at every rational but not everywhere I don't think such a sequence exists. 
I have a proof (given in class) that if the sequence is equicontinuous and converges point-wise (non-uniformly) for every rational on $[a,b]$ then it converges uniformly on all of $[a,b]$. 
I also have a proof (written by me) that if a sequence of continuous functions defined on $[a,b]$ converges point-wise for every rational on $[a,b]$, then it  converges.

EDIT: "uniformly" change to "pointwise"

on all of $[a,b]$. 
But I was told that equicontinuity is essential. I'm having hard time seeing why. Can someone give me a counterexample to demonstrate the need for equicontinuity?
My understanding is at the level of Rudin's principles [...] chapters 1-6 and most of chapter 7. 
 A: $f_n(x) = (-1)^n(x/\sqrt 2)^n$ converges pointwise to $0$ everywhere on $[0,\sqrt 2),$ hence at each rational in $[0,\sqrt 2],$ but does not converge at $\sqrt 2.$
A: The following example is about why for an arbitrary sequence of functions (without equicontinuity assumed) one cannot conclude uniform continuity from pointwise convergence at every rational.
On $[0,2]$ let $f(x)=0$ if $x\le \sqrt{2}$ and $1$ otherwise. Take each $f_n$ to be this $f.$ Note this $f$ is not continuous at $\sqrt{2}.$
I'll add to this if I can think up such a sequence of functions each of whih is continuous. 
NOTE I just saw user zhw.'s answer above which is a sequence of continuous functions on $[0,1]$ which converges at every point (rational or not) and yet the convergence is not uniform (and limit function discontinuous). So I won't look further.
A: Define the sequence of functions $(f_n)_n$ as follows: for any $n\geq 1$,


*

*$f_n(x) = 0$ for $x\in[0,\frac{1}{\sqrt{2}}-\frac{1}{10n}]\cup [\frac{1}{\sqrt{2}}+\frac{1}{10n},1]$;

*$f_n(\frac{1}{\sqrt{2}}) = n$;

*$f_n$ is piecewise affine and continuous;
so that $f_n$ is a continuous function looking like a very narrow triangle centered at $\frac{1}{\sqrt{2}}$.
Then $f_n(x) \xrightarrow[n\to\infty]{}0$ for any $x\in[0,1]\setminus\{\frac{1}{\sqrt{2}}\}$, and a fortiori at all rationals. But there is not pointwise convergence on $[0,1]$, since $f_n(\frac{1}{\sqrt{2}}) \xrightarrow[n\to\infty]{}\infty$.
